A) \[30{}^\circ \]
B) \[60{}^\circ \]
C) \[75{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: B
Solution :
[b] Radius of circular path of the particle is \[r=\frac{mv}{qB}=\frac{2\times {{10}^{-3}}\times 0.3}{{{10}^{-3}}\times 0.2}=3m\] C is centre of the circular path. Particle enters the field at A and leaves at B. \[\tan \theta =\frac{R}{r}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\] Deviation \[=2\theta =60{}^\circ .\] [A careful observation tells that direction of velocity at B is along the radius of the given circle].You need to login to perform this action.
You will be redirected in
3 sec